Problem 3

Question

Jacobians and Transformed Regions in the Plane. a. Solve the system $$ u=3 x+2 y, \quad v=x+4 y $$ for $x$ and $y$ in terms of $u$ and $v .$ Then find the value of the Jacobian $\partial(x, y) / \partial(u, v)$ b. Find the image under the trunsformation $u=3 x+2 y$ $v=x+4 y$ of the triangular region in the $x y$ -plane bounded by the $x$ -axis, the $y$ -axis, and the line $x+y=1$. Sketch the transformed region in the $u v$ -plane.

Step-by-Step Solution

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Answer

The Jacobian is $\frac{1}{10}$. The region transforms into a triangle in the $uv$-plane with vertices at (0,0), (2,4), and (5,1).

1Step 1: Expressing x and y in terms of u and v

First, we start by solving the given system of equations for $x$ and $y$ in terms of $u$ and $v$. We have:\[u=3x+2y\]\[v=x+4y\]To solve this, we can use substitution or elimination methods. Let's use elimination. Multiply the second equation by 3 and subtract the first equation:\[3(v) - (u) = 3(x + 4y) - (3x + 2y)\]\[3v - u = 10y\]\[y = \frac{3v - u}{10}\]Substitute $y$ back into one of the original equations (we'll use $u = 3x + 2y$):\[u = 3x + 2\left(\frac{3v - u}{10}\right)\]\[10u = 30x + 6v - 2u\]\[12u = 30x + 6v\]\[30x = 12u - 6v\]\[x = \frac{12u - 6v}{30}\]\[x = \frac{2u - v}{5}\]

2Step 2: Finding the Jacobian

To find the Jacobian of the transformation, we calculate the determinant of the Jacobian matrix formed by partial derivatives:The Jacobian matrix, $J$, is:\[J = \begin{bmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{bmatrix}\]From Step 1, $x = \frac{2u - v}{5}$ and $y = \frac{3v - u}{10}$. Therefore:\[\frac{\partial x}{\partial u} = \frac{2}{5}, \quad \frac{\partial x}{\partial v} = -\frac{1}{5}\]\[\frac{\partial y}{\partial u} = -\frac{1}{10}, \quad \frac{\partial y}{\partial v} = \frac{3}{10}\]The determinant of $J$ is:\[\left| \begin{matrix}\frac{2}{5} & -\frac{1}{5} \-\frac{1}{10} & \frac{3}{10}\end{matrix} \right|\]Compute the determinant:\[= \frac{2}{5} \cdot \frac{3}{10} - (-\frac{1}{5} \cdot -\frac{1}{10})\]\[= \frac{6}{50} - \frac{1}{50}\]\[= \frac{5}{50}\]\[= \frac{1}{10}\]

3Step 3: Find the image of the region

Now, let's transform the region defined by the lines in the $xy$-plane:1. $x = 0$ transforms to $u = 2y$, $v = 4y$.2. $y = 0$ transforms to $u = 3x$, $v = x$.3. $x + y = 1$ transforms to $u = 3x + 2(1-x) = 2x + 2$, $v = x + 4(1-x) = 4 - 3x$.These will form the boundaries of the image in the $uv$-plane.

4Step 4: Sketch the transformed region

From the transformations:- For $x = 0$, points along this line become a line through the origin with slope $\frac{v}{u} = \frac{4}{2} = 2$.- For $y = 0$, let $u = 3v$, this line also passes through the origin with slope $1/3$.- Transform $x + y = 1$ to get points between $(u, v) = (2, 4)$ and $(u, v) = (5, 1)$.The region transforms into a triangle in the $uv$-plane with vertices at the origin, $(2, 4)$, and $(5, 1)$.

Key Concepts

Linear transformationsPartial derivativesDeterminantsCoordinate transformations

Linear transformations

Linear transformations play a central role in understanding Jacobians. They provide a way to map one vector space onto another using a linear function. Consider the transformation given by the equations:

$u = 3x + 2y$
$v = x + 4y$

This transformation takes coordinates from the $xy$-plane and translates them into the $uv$-plane.
A central feature of linear transformations is that they preserve operations like vector addition and scalar multiplication.
This makes them predictable and easy to manipulate mathematically. They effectively "reshape" the coordinate plane.
This process can involve stretching, rotating, or translating the original shape.
In this problem, the transformation is illustrated through turning lines in one plane into lines in another, visualized as shifting the vertices of geometric shapes when transferred from one space to another.

Partial derivatives

Partial derivatives help us understand how functions change when one of the inputs is varied, and are crucial when discussing Jacobians.
Given the equations for our transformation$(u = 3x + 2y, v = x + 4y)$, we aim to express $x$ and $y$ in terms of $u$ and $v$, which leads us to partial derivatives.
For instance, by finding $\frac{\partial x}{\partial u}$ and $\frac{\partial x}{\partial v}$, we measure how $x$ changes as $u$ or $v$ changes, keeping the other constant.
This requires solving the initial linear system, allowing us to state:

$x = \frac{2u - v}{5}$
$y = \frac{3v - u}{10}$

By applying partial differentiation to these expressions, we understand the rate of change of each variable, which is integral in forming the Jacobian matrix.
These derivatives show us the sensitivity of each transformed coordinate with respect to changes in the original ones.

Determinants

When calculating the Jacobian, determinants appear as an essential tool.
For the transformation from $xy$-plane to $uv$-plane, the determinant of the Jacobian matrix tells us how the transformation affects area.The Jacobian matrix $J$ for our transformation consists of partial derivatives, forming a matrix:\[J = \begin{bmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{bmatrix}\]Where,

$\frac{\partial x}{\partial u} = \frac{2}{5}$, $\frac{\partial x}{\partial v} = -\frac{1}{5}$
$\frac{\partial y}{\partial u} = -\frac{1}{10}$, $\frac{\partial y}{\partial v} = \frac{3}{10}$

To find the determinant of $J$, compute:\[\text{det}(J) = \left(\frac{2}{5}\right) \left(\frac{3}{10}\right) - \left(-\frac{1}{5}\right) \left(-\frac{1}{10}\right) = \frac{1}{10}\]This result of $\frac{1}{10}$ indicates that the transformation scales areas by this factor, giving insight into how the region's dimensions change during mapping from one plane to another.

Coordinate transformations

Coordinate transformations describe how a geometric shape or region is modified by the mapping process.
For this exercise, we explored this using a triangle in the $xy$-plane, outlined by the axes and line $x + y = 1$.The transformation under $u = 3x + 2y$ and $v = x + 4y$ involves:

The line $x = 0$ becomes $u = 2y$ and $v = 4y$. This simplifies to a line through the origin with slope 2 in the $uv$-plane.
The line $y = 0$ turns into a line $u = 3v$, which passes through the origin with a slope of $\frac{1}{3}$.
Transforming $x + y = 1$ leads to endpoints in the transformation of $(u, v) = (2, 4)$ and $(5, 1)$.

Together, these result in a new triangular region in the $uv$-plane. It highlights how each original line morphs in space, showing critical comprehension of operational changes and their geometric consequences. This view into coordinate transformation is pivotal in studying more complex mappings across different mathematical problems and models.

Problem 2

Problem 3

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